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LeeCode from 0 ——58. Length of Last Word

n-1 循環 from num defined pty cte def str

58. Length of Last Word

Given a string s consists of upper/lower-case alphabets and empty space characters ‘ ‘, return the length of last word in the string.

If the last word does not exist, return 0.

Note: A word is defined as a character sequence consists of non-space characters only.

Example:

Input:
"Hello World" Output: 5
解題思路:
1)判斷是否以空格結尾,且以末端空格所占位數;
2)從末端非空格位開始,向前循環,非空格位加1,循環到新的空格位則跳出循環。
c++代碼如下:
class Solution {
public:
int lengthOfLastWord(string s) {
int n=s.length();
int sum=0;
int nums=0;
for(int j=n-1;j>=0;j--){
if(s[j]==‘ ‘)
nums++;
else
break;
}
for(int i=n-nums-1;i>=0;i--){

if(s[i]!=‘ ‘)
sum++;
else
break;
}
return sum;

}
};

LeeCode from 0 ——58. Length of Last Word